lesson 10: transformer performance and operation notes/lesson 10...lesson 10: transformer...

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Lesson 10: Transformer

Performance and Operation ET 332b Ac Motors, Generators and Power Systems

1 Lesson 10_et332b.pptx

Learning Objectives

Lesson 10_et332b.pptx 2

After this presentation you will be able to:

Define transformer voltage regulation and compute its

value

Convert impedance values into per unit or percent

values based on transformer rating or other given

values

Perform calculations using the per unit system

Convert per unit/percent values into circuit values

Compute transformer circuit parameters from test

values.

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Transformer Voltage Regulation


Definition: Difference between output voltage at no load and

output voltage at rated load divided by rated voltage

Where VR = rated output voltage

VNL = full load output voltage

%100V

VVVR%

R

RNL

Also given in "Per Unit" - fraction from 0 - 1.0

R

RNL

V

VVVR

Lower values of regulation are better. Indicates that there is less voltage drop

across the transformer.

Note: all

voltages are

magnitude only

Negative regulation is possible. Indicates voltage rise across transformer. ( due

to leading p.f. load)

Voltage Regulation Circuit Model


Voltage regulation found through calculation that uses the total winding

impedance

LSeqLSLSLS VZIE To compute %VR %100V

VEVR%

LS

LSNL

Where: VLS = rated low side voltage (switch closed)

ELS = no load low side voltage (switch open)

ILS = low side load current at specified P.F.

ZeqLS = total winding impedance referred to L.V. side

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Transformer Voltage Regulation


Example 10-1: A 500 kVA 7200 - 2400 V single-phase

transformer is operating at rated load with a power factor

of 0.82 lagging. The total winding resistance and reactance

values referred to the high voltage side are Req = 0.197 and

Xeq = 0.877 ohms. The load is operating in step-down mode.

Sketch the appropriate equivalent circuit and determine:

a) equivalent low side impedance

b) the no-load voltage, ELS

c) the voltage regulation at 0.82 lagging power factor

d) the voltage regulation at 0.95 leading power factor

Example 10-1 Solution (1)


a) Refer impedances to low voltage side of transformer

b) no-load secondary voltage

ELS= voltage required to supply rated power at rated voltage so..

Ans Ans

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Determine the phase angle on the current using the power factor

Lagging Fp means

negative angle

Ans



c) Compute the percent voltage regulation for 0.82 lagging power factor

Ans

d) Compute the percent voltage regulation for 0.95 leading power factor

Compute the no-load voltage

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No-load voltage is

smaller than Vs

Compute regulation

Negative %VR

indicates LC

resonance in

transformer/load

combination

Per Unit and Percent Impedance of

Transformers


Equivalent circuit calculation requires knowledge of turns ratio and

reference of impedance values from one side to other. Per Unit system

is normalization scheme that removes the effect of turns ratio from

power system calculations.

Transformer manufacturers list impedances for transformers using

Per Unit or Percent impedance method

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Per Unit and Percent Impedance of

Transformers


Per Unit impedance calculation needs base quantities

Per Unit Method

Define base power and voltage. Compute base Z and I from these

quantities. Divide actual impedances. voltages and currents by bases to

get per unit values

For transformers

ratedbase SS

ratedbase VV

Base power defined as the rated power of the

electrical device

Base voltage defined as the rated voltages of the

transformer

Per Unit Computation Method


Computing Per Unit (percent) bases

base

basebase

V

SI

base

basebase

I

VZ

base

2

basebase

S

VZ or

Resistance, reactance and impedances can now be divided by Zbase to

get per unit values based on transformer rated power and voltage.

Multiply per unit by 100 to get percent impedance.

base

actpu

Z

ZZ

base

actpu

Z

RR

base

actpu

Z

XX

Where: Zact = device impedance in ohms

Ract = device resistance in ohms

Xact = device reactance in ohms

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Per Unit Calculations


pupupu XjRZ 2

pu

2

pupu XRZ

pu

pu1

R

Xtan

Per Unit (percent) impedances and components also

add as vectors

Ohm’s law and all other circuit theorems are valid for p.u.

impedances, voltages, currents and power

Other p.u. quantities

base

actpu

base

actpu

V

VV

I

II

base

actpu

base

actpu

S

QQ

S

PP

Per Unit Values of a Transformer


Example 10-2: The equivalent circuit above is for a single phase 25

kVA 7200 - 240 volt transformer. The parameters have the following

values:

Rp = 1.40 W Xlp = 0.25 W Rs = 0.11 W Xls = 3.20 W

Rfe = 19,501 W Xm = 5011 W

Convert these values to per unit values based on the current and

voltage ratings of the primary and secondary of the transformer.

Draw the equivalent circuit with the per unit values labeled.

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Select Vbase=7200 V and Sbase=25 kVA for primary side

Divide all actual values located on primary side by Zbase

Ans Ans



Ans Ans

Convert secondary values to per unit

Compute p.u. values

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Refer the Xls and Rs to primary side and compute p.u. values using

primary Zbase

Same values as low-voltage side p.u. calculation. Per unit with

transformer voltage ratings as base voltages removes turns ratio

from all calculations

Transformer Ratios and Per Unit


When common power base is used, and voltage bases are defined as

transformer rated voltages, p.u. (percent) values are the same on both

side of transformer. This means that the ideal transformer can be

removed from schematic and calculations done in p.u.

To get actual values from p.u.

pubaseact III pubaseact VVV pubaseact ZZZ

pubaseact PSP pubaseact QSQ pubaseact SSS

Note: Rated voltage and current values are 1.0 p.u.

pupupu ZIV pu

pu

puZ

VI

pu

pu

puI

VZ

Ohm's Law holds in p.u. so....

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Per Unit Calculations


Example 10-3: A 50 kVA 7200-240 V single phase transformer has an equivalent series impedance of 2575° Ohms in terms of the high

voltage side. The transformer supplies a 45 kVA load with 0.89 lagging

power factor at rated voltage. Find:

a) Per unit Zeq with rated high-side voltage as the base voltage

b) Per unit Zeq with rated low-side voltage as the base voltage

c) The %VR using per unit values



a) Primary side values in p.u.

b) Secondary side values in p.u.

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c) Find the %VR



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Per Unit Circuit Analysis


Example 10-4: The circuit shown below has a base voltage of 240 V and

base power of 1500 VA. Find the P.U. and actual current in the circuit.

Example 10-4 solution

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Find actual value of current

Power System Application of Per Unit


Example: 10-5: A 250 kVA 2400 - 240 V transformer

with a 2.2% impedance was damaged by a zero impedance

short across its low voltage terminals. Assuming rated

voltage and an impedance angle of 75 degrees, find:

a) the actual short circuit current,

b) the required percent impedance of the new

transformer to limit the short circuit current to 25,000

amps.

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a) Find actual short circuit current

Isc is approximately 45.5 times rated

Secondary current



b) Find per unit impedance that limits ISC to 25,000 amps

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Transformer Losses and Efficiencies


Definition of efficiency

Where: Po = transformer output power

Pcore = transformer core losses (from Open Circuit

test)

IL = load current (primary or secondary)

Req = total equivalent coil resistance (referred to

primary or secondary (from Short Circuit test)

eq

2

Lcoreo

o

RIPP

P

Efficiencies range from 96 -99% for large power transformers

Transformer Testing-Open Circuit

Test


Open circuit test finds Rfe and Xm - core losses and magnetizing

reactance

Test Set-up and conditions

P = wattmeter

A = ammeter

V = voltmeter

Test performed on low voltage side with H.V. side open circuited.

Voc = rated low side voltage. Measure: Ioc, Pc and Voc

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Test


Circuit Model Formulas for Finding Rfe and XM

Parallel circuit so Voc applied

across both elements

fe

ocfeLS

I

VR

M

ocMLS

I

VX

c

2

ocfeLS

P

VR or

LS indicates that values determined on the low voltage side


Test


Values found from open circuit test:

Pc = core loss power

Voc = open circuit voltage

Ioc = open circuit current

oc

cfe

V

PI

2

fe

2

ocM III

c

2

ocfeLS

P

VR

Find the active part of the current from the power and voltage readings

Find reactive current and

compute the value of XM

M

ocMLS

I

VX

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Open-Circuit Test Example


Example 10-6: An Open circuit test is performed on the

240 V windings of a 7200-240 V power transformer. The

following data are recorded for the test Voc = 240 V Ioc

= 16.75 A Pc = 580 W. Calculate the exciting resistance

and reactance.

Transformer Testing-Short Circuit Test


Short circuit test finds the total winding resistance and leakage

reactance for both coils referred to the side on which the test is

performed (usually H.V.)

Test Set-up and Conditions

Short circuit L.V. Side. Adjust the source voltage on H.V side until

rated H.V. current flows. NOTE: START WITH SOURCE V AT

ZERO. Measure: Vsc, Isc and Psc

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Transformer Testing-Short Circuit Test


Circuit model for finding Req and Xeq.

Formulas for finding ReqHS and XeqHS

eqHSeqHSeqHS XjRZ

sc

sceqHS

I

VZ

2

sc

sceqHS

I

PR

2

eqHS

2

eqHSeqHS RZX

The values of R and X are

found using H.V. side values

and can be referred to the

L.V. side by dividing by a2

Short Circuit Test Example


Example 10-7: The test data for a 75 kVA 7200 - 480 V single phase

transformer are listed below:

Open-Circuit Test Short-Circuit Test

(Low side data) (High side data)

Voc = 480 V Vsc = 173.1 V

Ioc = 16.5 A Isc = 16.3 A

Poc = 558 W Psc = 1200 W

Determine:

a) the core resistance and reactance, the equivalent winding resistance

and reactance and draw the equivalent circuit of the transformer

b) the per unit values of the values found in part a.)

c) the efficiency of the transformer when operating at rated load and

0.85 lagging power factor.

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Compute parameters from S.C. test



Transformer schematic showing all parameters

b) Find the per unit values of all circuit components

On primary side

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Compute P.U. values using Zbase

On secondary side

c) Find the efficiency at rated load



Use primary side voltage

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d) Find the %VR

Core R and XM do not contribute to voltage drop. Use simplified model.



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END LESSON 10:

TRANSFORMER

PERFORMANCE AND

OPERATION

ET 332b Ac Motors, Generators and Power Systems


lesson 10: transformer performance and operation notes/lesson 10...lesson 10: transformer...

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